Integrals Test

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Integrals Test - 38

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Question 1
1 / -0

$$\int _{ 0 }^{ 1 }{ \dfrac { dx }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right) } } $$=

A
$$\dfrac { \pi }{ 4 } +\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$

B
$$\dfrac { \pi }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$

C
$$\dfrac { \pi }{ 4 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$

D
$$\dfrac { \pi }{ 3 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$

Solution
Question 2
1 / -0

$$\int \left( x ^ { 6 } + 7 x ^ { 5 } + 6 x ^ { 4 } + 5 x ^ { 3 } + 4 x ^ { 2 } + 3 x + 1 \right) e ^ { x } d x$$ equals

A
$$\sum _ { j = 0 } ^ { 6 } x ^ { j } e ^ { x } + c$$

B
$$\sum _ { j = 1 } ^ { 7 } x ^ { j } e ^ { x } + c$$

C
$$\sum _ { i = 1 } ^ { 6 } x ^ { i } e ^ { x } + c$$

D
$$\sum _ { i = 0 } ^ { 5 } x ^ { i } e ^ { x } + c$$

Solution

$$\displaystyle\int (x^6+7x^5+6x^4+5x^3+4x^2+3x+1)e^xdx$$
$$=\displaystyle\int (x^6+6x^5)e^xdx+\displaystyle\int (x^5+5x^4)e^xdx+\displaystyle\int (x^4+4x^3)e^xdx+\displaystyle\int (x^3+3x^2)e^xdx=\displaystyle\int (x^2+2x)e^xdx+\displaystyle\int (x+1)e^xdx$$.
$$\displaystyle\int [f(x)+f'(x)]e^xdx=f(x)e^x$$
Using this fact, we get
$$\displaystyle\int (x^6+6x^5)e^xdx=\displaystyle\int (x^6+(x^6)')e^xdx$$
$$=x^6\cdot e^xdx$$
Hence G.E$$=x^6e^x+x^5e^x+...…+xe^x+c$$
$$=\displaystyle\sum^6_{i=1}x^ie^x+c$$.
Question 3
1 / -0

Evaluate$$\displaystyle \int _ { 0 } ^ { a } \dfrac { x d x } { \sqrt { a ^ { 2 } + x ^ { 2 } } } $$

A
$$a ( \sqrt { 2 } - 1 )$$

B
$$a ( 1 - \sqrt { 2 } )$$

C
$$a ( 1 + \sqrt { 2 } )$$

D
$$2 a \sqrt { 3 }$$

Solution

$$\displaystyle \int _{ 0 }^{ a }{ \cfrac { xdx }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } } $$

Let $$t=\left( { a }^{ 2 }+{ x }^{ 2 } \right) \Rightarrow dt=2xdx$$

when $$x=0,t={ a }^{ 2 }\quad $$

when $$x=a,t=2{ a }^{ 2 }$$

$$\displaystyle =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ \cfrac { 2xdx }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } } =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ \cfrac { dt }{ \sqrt { t } } } =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ { t }^{ -1/2 }dt } $$

$$=\cfrac { 1 }{ 2 } \quad { \left[ \cfrac { { t }^{ -1/2+1 } }{ -\cfrac { 1 }{ 2 } +1 } \right] }_{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }=\cfrac { 1 }{ 2 } \quad { \left[ \cfrac { \sqrt { t } }{ \cfrac { 1 }{ 2 } } \right] }_{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }=\cfrac { 2 }{ 2 } \left[ \sqrt { 2{ a }^{ 2 } } -\sqrt { { a }^{ 2 } } \right] $$

$$=a\left( \sqrt { 2 } -1 \right) $$
Question 4
1 / -0

Evaluate $$\displaystyle \int _ { 0 } ^ { \infty } \frac { x ^ { 2 } + 1 } { x ^ { 4 } + 7 x ^ { 2 } + 1 } d x $$

A
$$\pi$$

B
$$\dfrac { \pi } { 2 }$$

C
$$\dfrac { \pi } { 3 }$$

D
$$\dfrac { \pi } { 6 }$$

Solution

$$\displaystyle\int_0^{\infty}\dfrac{x^2+1}{x^4+7x^2+1}dx$$

Let
$$I=\displaystyle\int_0^{\infty}\dfrac{x^2+1}{x^4+7x^2+1}dx$$

$$=\displaystyle\int_0^{\infty}\dfrac{1+\dfrac{1}{x^2}}{x^2+7+\dfrac{1}{x^2}}$$

$$=\displaystyle\int_0^{\infty}\dfrac{1+\dfrac{1}{x^2}}{x^2-2+\dfrac{1}{x^2}+9}$$

$$=\displaystyle\int_0^{\infty}\dfrac{1+\dfrac{1}{x^2}}{\left(x-\dfrac{1}{x}\right)^2+(3)^2}$$

Let $$x-\dfrac{1}{x}=t$$
$$\left(1+\dfrac{1}{x^2}\right)dx=dt$$

$$I=\displaystyle\int_0^{\infty}\dfrac{dt}{t^2+(3)^2}$$

$$I=\left[\dfrac{1}{3}\tan^{-1}\left(\dfrac{t}{3}\right)+c\right]^{\infty}_0$$

$$I=\left[\dfrac{1}{3}\tan^{-1}\left(\dfrac{x^2-1}{3x}\right)+c\right]^{\infty}_0$$

$$I=\dfrac{1}{3}\left[\dfrac{\pi}{2}-\left(\dfrac{-\pi}{2}\right)\right]$$

$$I=\dfrac{\pi}{3}$$
Question 5
1 / -0

The intercepts on x-axis made by tangents to the curve, $$\int _{ 0 }^{ x }{ \left| t \right| } dt,x\in R,$$ which are parallel to the line y=2x, are equal to:

A
$$_{ - }^{ + }{ 3 }$$

B
$$_{ - }^{ + }4$$

C
$$_{ - }^{ + }1$$

D
$$_{ - }^{ + }2$$

Solution
Question 6
1 / -0

$$\int _{ 0 }^{ \pi }{ \cfrac { { x }^{ 2 } }{ { \left( 1+sinx \right) }^{ 2 } } } dx$$ equals

A
$$\pi (\pi -2)$$

B
$$\pi ^2 (\pi -2)$$

C
$$\pi (2-\pi )$$

D
none of these
Question 7
1 / -0

The value of $$\int _{ -1 }^{ 1 }{ \dfrac { { cot }^{ -1 }x }{ \pi } } dx$$

A
1

B
2

C
3

D
0

Solution
Question 8
1 / -0

The value of the integral $$\displaystyle \int_{-\pi/2}^{\pi/2} \left(x^{2}+\log \dfrac{\pi-x}{\pi+x}\right) \cos x dx $$ is

A
$$0$$

B
$$\dfrac{\pi^{2}}{2}-4$$

C
$$\dfrac{\pi^{2}}{2}+4$$

D
$$\dfrac{\pi^{2}}{2}$$

Solution

$$I=\displaystyle \int_{-\pi/2}^{\pi/2}\left[x^2+log\left( \dfrac{\pi-x}{\pi+x}\right)\right]cosxdx$$
As,$$\displaystyle \int_{-a}^{a}f(x)dx=0,$$when$$f(-x)=-f(x)$$
$$\therefore I=\displaystyle \int_{-\pi/2}^{\pi/2}x^2cosxdx+0=2\displaystyle \int_{0}^{\pi/2}(x^2cosx)dx$$
$$=2\left\{ (x^2sinx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}(x^3cosx) \right\}$$
$$2\left[ \dfrac{\pi^2}{4}-2\left\{(-xcosx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}1.(-cosx)dx \right\}\right]$$
$$=2\left[ \dfrac{\pi^2}{4}-2(sinx)_{0}^{\pi/2}\right]=2\left[\dfrac{\pi^2}{4}-2\right]=\left( \dfrac{\pi^2}{2}-4\right)$$
Question 9
1 / -0

If $$\int \frac { x ^ { 2 } \tan ^ { - 1 } x } { 1 + x ^ { 2 } } d x = \tan ^ { - 1 } x - \frac { 1 } { 2 } \log \left( 1 + x ^ { 2 } \right) + f ( x ) + c$$ then $$f ( x ) =$$

A
$$- \frac { \tan ^ { - 1 } x } { 2 }$$

B
$$- \frac { 1 } { 2 } \left( \tan ^ { - 1 } x \right) ^ { 2 }$$

C
$$\frac { \tan ^ { - 1 } x } { 2 }$$

D
None of these

Solution

$$\\\int\>(\frac{(1+x^2-1)tan^{-1}x}{1+x^2})dx\\=\int\>\left(1-(\frac{1}{1+x^2})\right)tan^{-1}x\>dx\\=\int\>1.tan^{-1}dx-\int\>(\frac{tan^{-1}x}{1+x^2})dx\\Use\>ILATE\>in\>first\>part\>and\>assume\\tan^{-1}x=t\>for\>second\>integration\\=xtan^{-1}x-\int\>(\frac{1}{1+x^2})\times\>xdx-\int\>t\>dt\\=xtan^{-1}x-(\frac{1}{2})log|1+x^2|-(\frac{t^2}{2})+C\\=xtan^{-1}x-(\frac{1}{2})log|1+x^2|-(\frac{(tan^{-1}x)^2}{2})+C\\\therefore\>C=-(\frac{1}{2})(tan^{-1}x)^2$$
Question 10
1 / -0

$$\begin{matrix} lim \\ n\rightarrow \infty \end{matrix}\int _{ 0 }^{ 1 }{ \frac { { nx }^{ { n- }1 } }{ { 1+x }^{ 2 } } dx= } $$

A
0

B
1

C
2

D
$$\frac { 1 }{ 2 } $$

Solution